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"An expression can result in a value..."

I'm trying to improve my formal understanding of C++. One significant part
of that effort involves clarifying my understanding of the vocabulary used
to describe the language.

This is from the C++ Standard:

"[Note: Clause 5 defines the syntax, order of evaluation, and meaning of
expressions. An expression is a sequence of operators and operands that
specifies a computation. An expression can result in a value and
can cause side effects. ...]"

"...sequenc e of operators and operands that specifies a computation..." .
That means to me that an expression can be "executed". I am purposely
avoiding the term "evaluate" because that has connotations which I am not
sure apply to all expressions. Nonetheless, the notion of a computation
signifies to me that some kind of predetermined state change takes place at
the time the flow of control reaches the point in the compiled program
corresponding to the /expression/ appearing in the source code. So my
first question is:

How might we formulate a definition of "computatio n" which communicates the
intended meaning in the above quoted excerpt?

My second question involves the exact meaning of "An expression can result
in a value...". The concept of 'resulting in a value' suggests there is
some storage location holding said value at the instant at which the
execution ("evaluation " probably applies here) of the expression is
complete. I note that the Standard does not specifically state that an
expression can /return/ a value. It seems to me that the resulting value
of a function returning a type other than void will be the value returned.
But what about values assigned to non-const parameters passed by reference?
Are these to be considered side effects, resulting values, or both?

In C++ parlance, is the value resulting from `int x, y;/*...*/ x + y;'
considered a "return value"? Is there ever an instance in which
the 'resulting value' of an expression cannot properly be called
its 'return value'?
--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/
Dec 6 '06 #1
21 2445

Steven T. Hatton wrote:
I'm trying to improve my formal understanding of C++. One significant part
of that effort involves clarifying my understanding of the vocabulary used
to describe the language.

This is from the C++ Standard:

"[Note: Clause 5 defines the syntax, order of evaluation, and meaning of
expressions. An expression is a sequence of operators and operands that
specifies a computation. An expression can result in a value and
can cause side effects. ...]"

"...sequenc e of operators and operands that specifies a computation..." .
That means to me that an expression can be "executed". I am purposely
avoiding the term "evaluate" because that has connotations which I am not
sure apply to all expressions. Nonetheless, the notion of a computation
signifies to me that some kind of predetermined state change takes place at
the time the flow of control reaches the point in the compiled program
corresponding to the /expression/ appearing in the source code. So my
first question is:

How might we formulate a definition of "computatio n" which communicates the
intended meaning in the above quoted excerpt?

My second question involves the exact meaning of "An expression can result
in a value...". The concept of 'resulting in a value' suggests there is
some storage location holding said value at the instant at which the
execution ("evaluation " probably applies here) of the expression is
complete. I note that the Standard does not specifically state that an
expression can /return/ a value. It seems to me that the resulting value
of a function returning a type other than void will be the value returned.
But what about values assigned to non-const parameters passed by reference?
Are these to be considered side effects, resulting values, or both?

In C++ parlance, is the value resulting from `int x, y;/*...*/ x + y;'
considered a "return value"? Is there ever an instance in which
the 'resulting value' of an expression cannot properly be called
its 'return value'?
I'm not sure that I'm better off in my understanding than you, so what
I'm saying might be blatently obvious to you. (Or possibly wrong...)

But my understanding is that there's not a fundamental difference
between what you're thinking of a return value and what the standard
calls resulting in a value. 5+6; doesn't return a value, because it
doesn't *return* anything. "Return" is a statement about control flow.
(There might be a value attached to it, as in a function returning an
int, but the point is that it at least has a substantial part that is
control flow.)
But what about values assigned to non-const parameters passed by reference?
Are these to be considered side effects, resulting values, or both?
Are you saying that on the marked line:

int foo(xnt& x) { x=1; return 2; }
int bar() {
int a;
foo(a) + 5; // <= here
}

the change in value of a is considered a side effect, resulting value,
or both?

If so, that's definitely just a side effect. The resulting value of
'foo(a)' should be 2, and then of the whole expression statement, 7.

Evan

Dec 6 '06 #2
Steven T. Hatton <ch********@ger mania.supwrote:
>In C++ parlance, is the value resulting from `int x, y;/*...*/ x + y;'
considered a "return value"? Is there ever an instance in which
the 'resulting value' of an expression cannot properly be called
its 'return value'?
I only use "return value" for the value returned by a function.

Others may use the phrase differently.

Steve
Dec 6 '06 #3
Evan wrote:
>
Steven T. Hatton wrote:
>I'm trying to improve my formal understanding of C++. One significant
part of that effort involves clarifying my understanding of the
vocabulary used to describe the language.

This is from the C++ Standard:

"[Note: Clause 5 defines the syntax, order of evaluation, and meaning of
expressions. An expression is a sequence of operators and operands that
specifies a computation. An expression can result in a value and
can cause side effects. ...]"

"...sequenc e of operators and operands that specifies a computation..." .
That means to me that an expression can be "executed". I am purposely
avoiding the term "evaluate" because that has connotations which I am not
sure apply to all expressions. Nonetheless, the notion of a computation
signifies to me that some kind of predetermined state change takes place
at the time the flow of control reaches the point in the compiled program
correspondin g to the /expression/ appearing in the source code. So my
first question is:

How might we formulate a definition of "computatio n" which communicates
the intended meaning in the above quoted excerpt?

My second question involves the exact meaning of "An expression can
result
in a value...". The concept of 'resulting in a value' suggests there is
some storage location holding said value at the instant at which the
execution ("evaluation " probably applies here) of the expression is
complete. I note that the Standard does not specifically state that an
expression can /return/ a value. It seems to me that the resulting value
of a function returning a type other than void will be the value
returned. But what about values assigned to non-const parameters passed
by reference? Are these to be considered side effects, resulting values,
or both?

In C++ parlance, is the value resulting from `int x, y;/*...*/ x + y;'
considered a "return value"? Is there ever an instance in which
the 'resulting value' of an expression cannot properly be called
its 'return value'?

I'm not sure that I'm better off in my understanding than you, so what
I'm saying might be blatently obvious to you. (Or possibly wrong...)

But my understanding is that there's not a fundamental difference
between what you're thinking of a return value and what the standard
calls resulting in a value. 5+6; doesn't return a value, because it
doesn't *return* anything. "Return" is a statement about control flow.
(There might be a value attached to it, as in a function returning an
int, but the point is that it at least has a substantial part that is
control flow.)
It's been a long time since I went through the details of how an activation
stack works, but IIRC, the instructions resulting from compiling an
expression are put on the stack along with any automatic variables.
Variables which are visible from outside the code being executed have to be
referred to by their addresses. Any value "returned" by a function must be
placed in storage which can be accessed after the function returns. In
this context "returning" means the instruction pointer has reached the last
instruction of the function, and then assumed the address of the subsequent
instruction.

So, yes, there is a significant component of control flow involved in the
notion of "returning" . It seems reasonable to conclude that the address
range into which the return value (if any) will be placed must be
established prior to execution of the function producing the value.

I'm not sure how the address of a function used to initialize a pointer to
function relates to the storage location of a return value.

I do believe the notion of "returning" is applicable to arithmetic
expressions such as int x, y; x+y; I'm not sure how it might differ from
the branch and return that takes place with a user written function call.
One thing to keep in mind with C++ is that there is a notion of returning
references which makes me cautious about jumping to any conclusions about
what it means to have a return value. I'm also not sure what, if any,
differences exist between the execution time characteristics of a storage
location used for a return value and the storage location used for a
variable parameter.

My inclination is to think of 'return value' and 'resulting value' as
synonymous, but this is C++. Anything is possible.
>But what about values assigned to non-const parameters passed by
reference? Are these to be considered side effects, resulting values, or
both?

Are you saying that on the marked line:

int foo(xnt& x) { x=1; return 2; }
int bar() {
int a;
foo(a) + 5; // <= here
}

the change in value of a is considered a side effect, resulting value,
or both?

If so, that's definitely just a side effect. The resulting value of
'foo(a)' should be 2, and then of the whole expression statement, 7.
My inclination is to agree, but, as I said, this is C++.
--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/
Dec 6 '06 #4
It's been a long time since I went through the details of how an activation
stack works, but IIRC, the instructions resulting from compiling an
expression are put on the stack along with any automatic variables.
I'm not sure what you mean here. Instructions don't go on the stack.
(In fact, in an ideal world, the stack would be marked no execute.) Do
you mean the intermediate computation results?
Variables which are visible from outside the code being executed have to be
referred to by their addresses. Any value "returned" by a function must be
placed in storage which can be accessed after the function returns. In
this context "returning" means the instruction pointer has reached the last
instruction of the function, and then assumed the address of the subsequent
instruction.

So, yes, there is a significant component of control flow involved in the
notion of "returning" . It seems reasonable to conclude that the address
range into which the return value (if any) will be placed must be
established prior to execution of the function producing the value.
That depends on what you mean by "establishe d". If you mean known, then
yes. But if you mean "here's the return value spot", this isn't usually
how it's done. For instance, on x86 and the cdecl calling convention,
the return value spot is usually the same as either one of the local
variables or a spot for a saved register. With stdcall, I think it
shares a spot with one of the parameters. Either way, there's not a
dedicated "here's where the return value goes" location that's set
aside for that purpose. Then of course there are some architectures
(SPARC for instance) where standard calling conventions place the
return value in a register.

But what happens with the return value is secondary to my point, which
is that the standard is probably reserving "return" for the control
flow effects.

(Stuff like short-circuiting operators sorta throw a wrench into my
"expression s don't have control flow so don't return" argument, but
that's still the idea. Return is control flow from a function.)
I'm not sure how the address of a function used to initialize a pointer to
function relates to the storage location of a return value.
It doesn't.

I do believe the notion of "returning" is applicable to arithmetic
expressions such as int x, y; x+y;
I don't fault your use of 'return' there; I've used that myself.
("Postfix increment returns the value before the change.") But I think
that the standard is making that distinction, and why they don't use
return.
I'm not sure how it might differ from
the branch and return that takes place with a user written function call.
Because there's no control flow. (Again, short circuits throw a wrench
into this definition, but there's still no control flow away from a
function.)
One thing to keep in mind with C++ is that there is a notion of returning
references which makes me cautious about jumping to any conclusions about
what it means to have a return value.
What's special about references? It's just returning an address.
I'm also not sure what, if any,
differences exist between the execution time characteristics of a storage
location used for a return value and the storage location used for a
variable parameter.
Modulo architecture differences (SPARC puts these values into different
registers, etc.), not really anything. Both get placed on the stack in
the current activation record. Again in x86, the caller pushes the
function onto the stack, the callee pushes the return value onto the
stack, and the caller pops the return value off. (Who pops the
arguments depends on the calling convention. In cdecl, the caller pops,
while in stdcall the callee pops. There's also fastcall which puts some
parms in registers.)

Evan
>
My inclination is to think of 'return value' and 'resulting value' as
synonymous, but this is C++. Anything is possible.
But what about values assigned to non-const parameters passed by
reference? Are these to be considered side effects, resulting values, or
both?
Are you saying that on the marked line:

int foo(xnt& x) { x=1; return 2; }
int bar() {
int a;
foo(a) + 5; // <= here
}

the change in value of a is considered a side effect, resulting value,
or both?

If so, that's definitely just a side effect. The resulting value of
'foo(a)' should be 2, and then of the whole expression statement, 7.

My inclination is to agree, but, as I said, this is C++.
--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/
Dec 6 '06 #5
Steven T. Hatton wrote:
I'm trying to improve my formal understanding of C++. One significant
part of that effort involves clarifying my understanding of the vocabulary
used to describe the language.

This is from the C++ Standard:

"[Note: Clause 5 defines the syntax, order of evaluation, and meaning of
expressions. An expression is a sequence of operators and operands that
specifies a computation. An expression can result in a value and
can cause side effects. ...]"

"...sequenc e of operators and operands that specifies a computation..." .
That means to me that an expression can be "executed". I am purposely
avoiding the term "evaluate" because that has connotations which I am not
sure apply to all expressions.
The standard uses the term "evaluate".
Nonetheless, the notion of a computation
signifies to me that some kind of predetermined state change takes place
at the time the flow of control reaches the point in the compiled program
corresponding to the /expression/ appearing in the source code. So my
first question is:

How might we formulate a definition of "computatio n" which communicates
the intended meaning in the above quoted excerpt?
The word "computatio n" occurs about four times in the standard and is
clearly used loosely. The closest you get is model of the the abstract
machine [1.9].
My second question involves the exact meaning of "An expression can result
in a value...". The concept of 'resulting in a value' suggests there is
some storage location holding said value at the instant at which the
execution ("evaluation " probably applies here) of the expression is
complete.
Nope, it does not suggest anything like that.
I note that the Standard does not specifically state that an
expression can /return/ a value. It seems to me that the resulting value
of a function returning a type other than void will be the value returned.
Correct.
But what about values assigned to non-const parameters passed by
reference? Are these to be considered side effects, resulting values, or
both?
No. But that answer relies on a very pedantic reading of the question.

In C++ parlance, is the value resulting from `int x, y;/*...*/ x + y;'
considered a "return value"? Is there ever an instance in which
the 'resulting value' of an expression cannot properly be called
its 'return value'?
As far as I know, the standard uses the term return value only for
functions, operator functions, and conversions. Built in operators are none
of those.
Best

Kai-Uwe Bux
Dec 6 '06 #6
Evan wrote:
>It's been a long time since I went through the details of how an
activation stack works, but IIRC, the instructions resulting from
compiling an expression are put on the stack along with any automatic
variables.

I'm not sure what you mean here. Instructions don't go on the stack.
(In fact, in an ideal world, the stack would be marked no execute.) Do
you mean the intermediate computation results?
You are correct. The only thing that typically goes on the stack is the
return address. I was mixing paradigms. In Mathematica operators and
operands are treated on a peer. (more or less). For example, a recursive
function such as this simple implementation of factorial will push
everything onto a stack and then evaluate it as the stack pops.

f[1] = 1;
f[n_] := f[n - 1]n
Trace[f[4]]

{f[4],
f[4 - 1] 4,
{
{4 - 1, 3},
f[3],
f[3 - 1] 3,
{
{3 - 1, 2},
f[2], f[2 - 1] 2,
{
{2 - 1, 1},
f[1], 1
},
1 2, 2
},
2 3, 6
},
6 4, 24
}
I am a bit confused about what goes on with inline functions. I had been
given to believe they go "on the stack" but upon further investigation,
that appears incorrect.
That depends on what you mean by "establishe d". If you mean known, then
yes. But if you mean "here's the return value spot", this isn't usually
how it's done. For instance, on x86 and the cdecl calling convention,
the return value spot is usually the same as either one of the local
variables or a spot for a saved register. With stdcall, I think it
shares a spot with one of the parameters. Either way, there's not a
dedicated "here's where the return value goes" location that's set
aside for that purpose. Then of course there are some architectures
(SPARC for instance) where standard calling conventions place the
return value in a register.
What I really meant is that there is some kind of activation record format,
and the address of the return value is established as an offset relative to
the starting address of that record.
But what happens with the return value is secondary to my point, which
is that the standard is probably reserving "return" for the control
flow effects.

(Stuff like short-circuiting operators sorta throw a wrench into my
"expression s don't have control flow so don't return" argument, but
that's still the idea. Return is control flow from a function.)
They use the term "returning" when talking about functions returning values.
They also use terminology talking about the "type" of an expression, which
I am inclined to understand as the "type" of the value it returns. I will
have to look more closely at the discussion of functions before I draw any
hard and fast conclusions.
>I do believe the notion of "returning" is applicable to arithmetic
expressions such as int x, y; x+y;

I don't fault your use of 'return' there; I've used that myself.
("Postfix increment returns the value before the change.") But I think
that the standard is making that distinction, and why they don't use
return.
I have not seen the term used in the context of arithmetic operations
performed on builtin types. I'm not sure what the significance of that is.

>One thing to keep in mind with C++ is that there is a notion of returning
references which makes me cautious about jumping to any conclusions about
what it means to have a return value.

What's special about references?
I'm not sure. I'm just being cautious.
It's just returning an address.
My understanding is that references don't really exist at runtime in the
same way that pointers, for example, exist.
> I'm also not sure what, if any,
differences exist between the execution time characteristics of a storage
location used for a return value and the storage location used for a
variable parameter.

Modulo architecture differences (SPARC puts these values into different
registers, etc.), not really anything. Both get placed on the stack in
the current activation record. Again in x86, the caller pushes the
function onto the stack, the callee pushes the return value onto the
stack, and the caller pops the return value off. (Who pops the
arguments depends on the calling convention. In cdecl, the caller pops,
while in stdcall the callee pops. There's also fastcall which puts some
parms in registers.)
Now I'm confused again. What exactly does it mean to push the function onto
the stack? Do you mean the activation record of the function? I would
expect that to hold the return value location. How the address of that
return value is communicated to the caller is not clear. But this really
goes beyond what I am trying to address right now.
--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/
Dec 6 '06 #7
Kai-Uwe Bux wrote:
Steven T. Hatton wrote:
>I'm trying to improve my formal understanding of C++. One significant
part of that effort involves clarifying my understanding of the
vocabulary used to describe the language.

This is from the C++ Standard:

"[Note: Clause 5 defines the syntax, order of evaluation, and meaning of
expressions. An expression is a sequence of operators and operands that
specifies a computation. An expression can result in a value and
can cause side effects. ...]"

"...sequenc e of operators and operands that specifies a computation..." .
That means to me that an expression can be "executed". I am purposely
avoiding the term "evaluate" because that has connotations which I am not
sure apply to all expressions.

The standard uses the term "evaluate".
I believe the term evaluate should be reserved for actions which actually
result in a value. Of course we can then haggle over whether a value which
is neither preserved nor ever used is actually the product of an
evaluation, since we never really "find the value". One could also argue
that all state changes are changes in value, so it would be hard for me to
argue very forcefully for my usage. I prefer, however, to refrain from
using the term "evaluate" when talking about dereferencing a pointer, for
example.
>My second question involves the exact meaning of "An expression can
result
in a value...". The concept of 'resulting in a value' suggests there is
some storage location holding said value at the instant at which the
execution ("evaluation " probably applies here) of the expression is
complete.

Nope, it does not suggest anything like that.
Then what is a value? And what does it mean to "result in a value"?
>But what about values assigned to non-const parameters passed by
reference? Are these to be considered side effects, resulting values, or
both?

No. But that answer relies on a very pedantic reading of the question.
I would have considered the modification of a variable parameter to be a
side effect. Do you not agree with that?
>In C++ parlance, is the value resulting from `int x, y;/*...*/ x + y;'
considered a "return value"? Is there ever an instance in which
the 'resulting value' of an expression cannot properly be called
its 'return value'?

As far as I know, the standard uses the term return value only for
functions, operator functions, and conversions. Built in operators are
none of those.
That's how it looks to me. I wonder if that is a conscious distinction or
simply the way things happened to turn out.
--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/
Dec 6 '06 #8
Steven T. Hatton wrote:
Kai-Uwe Bux wrote:
>Steven T. Hatton wrote:
>>I'm trying to improve my formal understanding of C++. One significant
part of that effort involves clarifying my understanding of the
vocabulary used to describe the language.

This is from the C++ Standard:

"[Note: Clause 5 defines the syntax, order of evaluation, and meaning of
expressions . An expression is a sequence of operators and operands that
specifies a computation. An expression can result in a value and
can cause side effects. ...]"

"...sequenc e of operators and operands that specifies a computation..." .
That means to me that an expression can be "executed". I am purposely
avoiding the term "evaluate" because that has connotations which I am
not sure apply to all expressions.

The standard uses the term "evaluate".

I believe the term evaluate should be reserved for actions which actually
result in a value. Of course we can then haggle over whether a value
which is neither preserved nor ever used is actually the product of an
evaluation, since we never really "find the value". One could also argue
that all state changes are changes in value, so it would be hard for me to
argue very forcefully for my usage. I prefer, however, to refrain from
using the term "evaluate" when talking about dereferencing a pointer, for
example.
>>My second question involves the exact meaning of "An expression can
result
in a value...". The concept of 'resulting in a value' suggests there is
some storage location holding said value at the instant at which the
execution ("evaluation " probably applies here) of the expression is
complete.

Nope, it does not suggest anything like that.

Then what is a value?
A value is an lvalue or an rvalue. For rvalues, there is no guarantee that
they are physically represented in memory.
And what does it mean to "result in a value"?
It means that the type of the expression is not void and the evaluation of
the expression does not invoke undefined behavior.

>>But what about values assigned to non-const parameters passed by
reference? Are these to be considered side effects, resulting values, or
both?

No. But that answer relies on a very pedantic reading of the question.

I would have considered the modification of a variable parameter to be a
side effect. Do you not agree with that?
I do agree with that, but the modification of the parameter is not the same
as the value that the parameter gets. The value maybe the number 5. The
number 5, by and in itself, is not a side effect; the side effect is that a
certain location in memory now will have 5 stored in it. If that memory
location was different, say because you passed a different parameter, the
side effect would be different even though the value might still be 5.
Since the value did not change but the side effect did change, the side
effect and the value cannot be identical.

>>In C++ parlance, is the value resulting from `int x, y;/*...*/ x + y;'
considered a "return value"? Is there ever an instance in which
the 'resulting value' of an expression cannot properly be called
its 'return value'?

As far as I know, the standard uses the term return value only for
functions, operator functions, and conversions. Built in operators are
none of those.

That's how it looks to me. I wonder if that is a conscious distinction or
simply the way things happened to turn out.
It is somewhat consistent: return values stem from return statements. I
would not read too much into it.
Best

Kai-Uwe Bux
Dec 6 '06 #9
Kai-Uwe Bux wrote:
Steven T. Hatton wrote:
>Kai-Uwe Bux wrote:
Then what is a value?

A value is an lvalue or an rvalue. For rvalues, there is no guarantee that
they are physically represented in memory.
That really doesn't answer the question. I'm not coming up with an example
of an rvalue who's value doesn't exist, at some time, in memory. I believe
there may be instances of such entities but nothing is coming to mind.
Nonetheless, why should such an entity be considered a value?
>And what does it mean to "result in a value"?

It means that the type of the expression is not void and the evaluation of
the expression does not invoke undefined behavior.
Does it really? Suppose there is some block of code which the compiler can
determine will not impact the observable result of running the program as
required by the Standard. The compiler can omit the code from the
executable program. Is it really meaningful to say that, given such a
circumstance, the expression was evaluated and a value resulted?
>
>>>But what about values assigned to non-const parameters passed by
reference? Are these to be considered side effects, resulting values,
or both?

No. But that answer relies on a very pedantic reading of the question.

I would have considered the modification of a variable parameter to be a
side effect. Do you not agree with that?

I do agree with that, but the modification of the parameter is not the
same as the value that the parameter gets. The value maybe the number 5.
The number 5, by and in itself, is not a side effect; the side effect is
that a certain location in memory now will have 5 stored in it. If that
memory location was different, say because you passed a different
parameter, the side effect would be different even though the value might
still be 5. Since the value did not change but the side effect did change,
the side effect and the value cannot be identical.
My statement predicated that the value was assigned to a particular
non-const parameter.
>That's how it looks to me. I wonder if that is a conscious distinction
or simply the way things happened to turn out.

It is somewhat consistent: return values stem from return statements. I
would not read too much into it.
I will never find a nuance in C++ that I do not hold suspect.
--
NOUN:1. Money or property bequeathed to another by will. 2. Something handed
down from an ancestor or a predecessor or from the past: a legacy of
religious freedom. ETYMOLOGY: MidE legacie, office of a deputy, from OF,
from ML legatia, from L legare, to depute, bequeath. www.bartleby.com/61/
Dec 6 '06 #10

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